How to print a list element by element vertically? Racket

Question

Im a noob in Racket, and I am trying to figure out how to print the elements of one list vertically, thus:

(printv '(1 2 3 4))

1

2

3

4

T

I try with if and cond, thus:

(define (delete x lst)
  (cond [(member x lst)(append(reverse(rest(member x(reverse lst))))(rest(member x lst)))]
        [else "Error"]
  ))


    (define (printv lst)
      (cond [(< 0 (length lst)) ((printf "~a ~%" (first lst))(printv (delete (first lst) lst)))]
            [else "T"]))

But I got this error in the moment the list becomes empty

application: not a procedure;
 expected a procedure that can be applied to arguments
  given: #<void>
  arguments...:

Help is appreciated

Answer 1

You should use built-in procedures to make your life easier. And why do you want to delete the elements? just iterate over the list. Try this:

(define (printv lst)
  (for-each displayln lst)
  (displayln "T"))

Answer 2

Just removing the extra parentheses fixes your problem:

(define (printv lst)
  (cond [(< 0 (length lst)) 
         (printf "~a ~%" (first lst))
         (printv (delete (first lst) lst))]
        [else "T"]))

Some hints:

Doing length at each iteration makes makes your function exponential time. Eg. doubling the list to display squares the time it takes. A beter approach to check if a list is empty is using (empty? lst)

How you get the rest of the list using delete is perhaps the most complicated I've seen. Since you know you are removing the first element you can just replace it with (rest lst) .

Just printing with a linefeed has a function: displayln

All combined:

(define (printv lst)
  (cond [(empty? lst) "T"]
        [else 
         (displayln (first lst))
         (printv (rest lst))]))

Pure Scheme version:

(define (printv lst)
  (cond [(null? lst) "T"]
        [else 
         (display (car lst))
         (newline)
         (printv (cdr lst))]))